(a)  notice that 21 mm is one standard deviation of 3 mm from the mean of 18 mm.

     Use the empirical rule that says that 68% of the data falls within one standard deviation
     from the mean.

     In this case, it means that half of 68%, i.e. 68/2 = 34% falls between 18 mm and 21 mm.
     Hence, 50% - 34% = 16% of data will exceed 21 mm.

     It is the ANSWER to (a).



(b)  To answer (b), you need to find the area under the normal curve 
     from z-score z1 = 17 mm to z-score z2 = 20 mm.

     Use the standard function normcdf of the regular calculator TI-83/84

                     z1  z2  mean  SD    <<<---===  formatting pattern.
         P = normcdf(17, 20, 18,   3)

     You will get the value  P = 0.3781  (rounded).

     So, the answer to (b) is  0.3781  (rounded).


     Alternatively, you may use an online calculator 
         https://onlinestatbook.com/2/calculators/normal_dist.html

     which is free of charge and has extremely convenient interface.



(c)  To answer (c), you need to find the z-score under the normal curve 
     such that the area of the normal curve on the right of z1 is 0.15.

     Use the standard function invNorm of the regular calculator TI-83/84

                     area  mean  SD    <<<---===  formatting pattern.
         z = invNorm(0.85, 18,   3)

     You will get the value  z = 21.1093  (rounded).

     So, the answer to (c) is  21.1093 mm  (rounded).


     Alternatively, you may use an online calculator 
         https://onlinestatbook.com/2/calculators/inverse_normal_dist.html

     which is free of charge and has extremely convenient interface.